Question

Researchers are interested in the relationship between cigarette smoking and lung cancer. Suppose an adult male is randomly selected from a particular population. The following table shows the probabilities of some events related to this chance process:

(a) Find the probability that the individual gets cancer given that he is a smoker. Show your work.

(b) Find the probability that the individual smokes or gets cancer. Show your work.

(c) Two adult males are selected at random. Find the probability that at least one of the two gets cancer. Show your work.

Short answer

Part (a) The probability is $0.32$

Part (b) The probability is $0.29$

Part (c) The probability is$0.2256$

Step-by-step solution

Part (a) Step 1. Given

The table is:

Part (a) Step 2. Concept used

Probability=$\frac{Favourableoutcomes}{Totaloutcomes}$

Part (a) Step 3. Calculation

If a person smokes, the probability of acquiring cancer is computed as follows:

$$\begin{gathered} P(Cancer|Smoker)=P(CancerandSmoker)/P(Smoker) \\ =0.08/0.25 \\ =0.32 \end{gathered}$$

As a result, the probability is $0.32$

Part (b) Step 1. Calculation

The likelihood that an individual will develop cancer or become a smoker is determined as follows:

$$\begin{gathered} P(SmokerorCancer)=1-P(Notcancernorsmoker) \\ =1-0.71 \\ =0.29 \end{gathered}$$

As a result, the probability is $0.29$

Part (c) Step 1. Calculation

The likelihood that at least one of the males will develop cancer is computed as follows:

$$\begin{gathered} P\left(Cancer\right)=P(SmokerorCancer)+P(SmokerorCancer)-P\left(Smoker\right) \\ =0.29+0.08-0.14 \\ =0.12 \end{gathered}$$

Now,

$$\begin{gathered} P(Atleastoncegetscancer)=1-P(2notcancer) \\ =1-(0.88)(0.88) \\ =1-0.7744 \\ =0.2256 \end{gathered}$$

As a result, the probability is $0.2256$